Mixing
Direct sum in physics
$begingroup$
I know this has been asked before but the questions tend to be specific to particular examples rather than the general case.
I still am not entirely sure when one needs a direct sum in physics. For example, on Wikipedia, it is said that
the $xy$-plane, a two-dimensional vector space, can be thought of as the direct sum of two one-dimensional vector spaces, namely the $x$ and $y$ axes. In this direct sum, the $x$ and $y$ axes intersect only at the origin (the zero vector). Addition is defined coordinate-wise, that is $(x_1 , y_1 ) + ( x_2 , y_2 ) = ( x_1 + x_2 , y_1 + y_2 )$ which is the same as vector addition.
Yet the dimension of the direct sum of two spaces is the sum of the dimension of the individual spaces. Clearly, the $xy$-plane is of dimension $mathbb{R}^2$ not $2mathbb{R}$.
I also looked at https://suchideas.com/articles/maths/math-phys/tensor-direct-products-vs-direct-sums/ which seems to have a nice table but again it's not clear why the direct sum corresponds to
two alternative groups of possibilities
In particular, their example of $w = 5u^{(1)} - 2v^{(5)}$ seems to be exactly the opposite of two exclusive groups as it combines an element from $U$ and another element from $V$.
Could anyone provide a clearer understanding with physical examples of direct sums in physics?
vectors linear-algebra
asked Jan 28 at 16:23
user1936752user1936752
5841515
$endgroup$
migrated from physics.stackexchange.com Jan 28 at 21:00
This question came from our site for active researchers, academics and students of physics.
add a comment |
$begingroup$
I know this has been asked before but the questions tend to be specific to particular examples rather than the general case.
I still am not entirely sure when one needs a direct sum in physics. For example, on Wikipedia, it is said that
the $xy$-plane, a two-dimensional vector space, can be thought of as the direct sum of two one-dimensional vector spaces, namely the $x$ and $y$ axes. In this direct sum, the $x$ and $y$ axes intersect only at the origin (the zero vector). Addition is defined coordinate-wise, that is $(x_1 , y_1 ) + ( x_2 , y_2 ) = ( x_1 + x_2 , y_1 + y_2 )$ which is the same as vector addition.
Yet the dimension of the direct sum of two spaces is the sum of the dimension of the individual spaces. Clearly, the $xy$-plane is of dimension $mathbb{R}^2$ not $2mathbb{R}$.
I also looked at https://suchideas.com/articles/maths/math-phys/tensor-direct-products-vs-direct-sums/ which seems to have a nice table but again it's not clear why the direct sum corresponds to
two alternative groups of possibilities
In particular, their example of $w = 5u^{(1)} - 2v^{(5)}$ seems to be exactly the opposite of two exclusive groups as it combines an element from $U$ and another element from $V$.
Could anyone provide a clearer understanding with physical examples of direct sums in physics?
vectors linear-algebra
asked Jan 28 at 16:23
user1936752user1936752
5841515
$endgroup$
migrated from physics.stackexchange.com Jan 28 at 21:00
This question came from our site for active researchers, academics and students of physics.
4
$begingroup$
If you are adding dimensions then you get $1+1=2$. I don't think I understand the issue. So you go from two spaces on $mathbb{R}$ (each dimension $1$) to the space on $mathbb{R}^2$ (dimension $2$).
$endgroup$
– Aaron Stevens
Jan 28 at 17:11
1
$begingroup$
Would Mathematics be a better home for this question.
$endgroup$
– Qmechanic
Jan 28 at 17:13
add a comment |
$begingroup$
I know this has been asked before but the questions tend to be specific to particular examples rather than the general case.
I still am not entirely sure when one needs a direct sum in physics. For example, on Wikipedia, it is said that
the $xy$-plane, a two-dimensional vector space, can be thought of as the direct sum of two one-dimensional vector spaces, namely the $x$ and $y$ axes. In this direct sum, the $x$ and $y$ axes intersect only at the origin (the zero vector). Addition is defined coordinate-wise, that is $(x_1 , y_1 ) + ( x_2 , y_2 ) = ( x_1 + x_2 , y_1 + y_2 )$ which is the same as vector addition.
Yet the dimension of the direct sum of two spaces is the sum of the dimension of the individual spaces. Clearly, the $xy$-plane is of dimension $mathbb{R}^2$ not $2mathbb{R}$.
I also looked at https://suchideas.com/articles/maths/math-phys/tensor-direct-products-vs-direct-sums/ which seems to have a nice table but again it's not clear why the direct sum corresponds to
two alternative groups of possibilities
In particular, their example of $w = 5u^{(1)} - 2v^{(5)}$ seems to be exactly the opposite of two exclusive groups as it combines an element from $U$ and another element from $V$.
Could anyone provide a clearer understanding with physical examples of direct sums in physics?
vectors linear-algebra
asked Jan 28 at 16:23
user1936752user1936752
5841515
$endgroup$
I know this has been asked before but the questions tend to be specific to particular examples rather than the general case.
I still am not entirely sure when one needs a direct sum in physics. For example, on Wikipedia, it is said that
the $xy$-plane, a two-dimensional vector space, can be thought of as the direct sum of two one-dimensional vector spaces, namely the $x$ and $y$ axes. In this direct sum, the $x$ and $y$ axes intersect only at the origin (the zero vector). Addition is defined coordinate-wise, that is $(x_1 , y_1 ) + ( x_2 , y_2 ) = ( x_1 + x_2 , y_1 + y_2 )$ which is the same as vector addition.
Yet the dimension of the direct sum of two spaces is the sum of the dimension of the individual spaces. Clearly, the $xy$-plane is of dimension $mathbb{R}^2$ not $2mathbb{R}$.
I also looked at https://suchideas.com/articles/maths/math-phys/tensor-direct-products-vs-direct-sums/ which seems to have a nice table but again it's not clear why the direct sum corresponds to
two alternative groups of possibilities
In particular, their example of $w = 5u^{(1)} - 2v^{(5)}$ seems to be exactly the opposite of two exclusive groups as it combines an element from $U$ and another element from $V$.
Could anyone provide a clearer understanding with physical examples of direct sums in physics?
vectors linear-algebra
vectors linear-algebra
asked Jan 28 at 16:23
user1936752user1936752
5841515
asked Jan 28 at 16:23
user1936752user1936752
5841515
asked Jan 28 at 16:23
user1936752user1936752
5841515
asked Jan 28 at 16:23
user1936752user1936752
5841515
asked Jan 28 at 16:23
user1936752user1936752
5841515
5841515
migrated from physics.stackexchange.com Jan 28 at 21:00
This question came from our site for active researchers, academics and students of physics.
migrated from physics.stackexchange.com Jan 28 at 21:00
This question came from our site for active researchers, academics and students of physics.
4
$begingroup$
If you are adding dimensions then you get $1+1=2$. I don't think I understand the issue. So you go from two spaces on $mathbb{R}$ (each dimension $1$) to the space on $mathbb{R}^2$ (dimension $2$).
$endgroup$
– Aaron Stevens
Jan 28 at 17:11
1
$begingroup$
Would Mathematics be a better home for this question.
$endgroup$
– Qmechanic
Jan 28 at 17:13
add a comment |
4
$begingroup$
If you are adding dimensions then you get $1+1=2$. I don't think I understand the issue. So you go from two spaces on $mathbb{R}$ (each dimension $1$) to the space on $mathbb{R}^2$ (dimension $2$).
$endgroup$
– Aaron Stevens
Jan 28 at 17:11
1
$begingroup$
Would Mathematics be a better home for this question.
$endgroup$
– Qmechanic
Jan 28 at 17:13
4
4
$begingroup$
If you are adding dimensions then you get $1+1=2$. I don't think I understand the issue. So you go from two spaces on $mathbb{R}$ (each dimension $1$) to the space on $mathbb{R}^2$ (dimension $2$).
$endgroup$
– Aaron Stevens
Jan 28 at 17:11
$begingroup$
If you are adding dimensions then you get $1+1=2$. I don't think I understand the issue. So you go from two spaces on $mathbb{R}$ (each dimension $1$) to the space on $mathbb{R}^2$ (dimension $2$).
$endgroup$
– Aaron Stevens
Jan 28 at 17:11
1
1
$begingroup$
Would Mathematics be a better home for this question.
$endgroup$
– Qmechanic
Jan 28 at 17:13
$begingroup$
Would Mathematics be a better home for this question.
$endgroup$
– Qmechanic
Jan 28 at 17:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The x-y plane is of dimension 2. While each line is of dimension 1. 1+1=2.
Let me tell you a bit more about direct sums. Given a vector space $U$ and two subspaces $V$ and $W$, we say $U$ is the direct sum of $V$ and $W$ (and write $U=Voplus W$) if the only vector which $V$ and $W$ share is $0$, i.e. $Vcap W={0}$, and every vector in $U$ can be written as a sum of a vector in $V$ and a vector in $W$, i.e. $U=V+W$. This notion of direct sum is called inner direct sum. It is very visual and easy to understand. For example, consider the $x$-axis $V:=text{span}{(1,0)}={(x,0)inmathbb{R}^2}$ and the $y$-axis $W:=text{span}{(0,1)}={(0,y)inmathbb{R}^2}$ in the plane. We then quickly check $U:=mathbb{R}^2=Voplus W$. Notice however that, unless we add more structure, the direct sum does not encode the orthogonality of the $x$ and $y$ axes. Indeed, if one considers the diagonal $Z:=text{span}{(1,1)}={(x,x)inmathbb{R}^2}$ we also have $U=Voplus Z$ and $U=Zoplus W$.
Now, suppose you have two (unrelated as far as we care) vector spaces $V$ and $W$. We may ask if there is some bigger vector space $U$ which contains $V$ and $W$ and $U=Voplus W$. This is always possible and the construction is called the outer direct sum. We define the space $Voplus W$ as the set of all tuples $(v,w)$ where $vin V$ and $win W$. This forms a vector space with the addition $(v_1,w_1)+(v_2,w_2):=(v_1+v_2,w_1+w_2)$ and scalar multiplication $k(v,w):=(kv,kw)$. We may then consider the spaces $tilde{V}:={(v,0)in Uoplus W}$ and $tilde{W}:={(0,w)in Uoplus W}$. $V$ and $W$ are isomorphic to $tilde{V}$ and $tilde{W}$ as vector spaces respectively. Both are subspaces of $Voplus W$ and in fact $Voplus W=tilde{V}oplustilde{W}$. In the left of this equation we have the outer direct sum just defined while in the second we have the inner direct sum defined in the previous paragraph. It is thus costumary to identify $V$ and $W$ with $tilde{V}$ and $tilde{W}$ repectively. Moreover, we use the notation $v+w:=(v,w)$ for all $vin V$ and $win W$. Then both notions of inner product are virtually undistinguishable and most authors don't care to distinguish between them. Notice that with the outer notion of direct sum we truly have $mathbb{R}^2=mathbb{R}oplusmathbb{R}$.
In physics the direct sum pops up whenever vector spaces do. In classical mechanics, for example, the position of a particle is described by the vector space (well, not really but let us avoid the problem of an origin for the moment) $V=mathbb{R}^3$. Now consider we have a second particle whose position is described by the vector space $W=mathbb{R}^3$. Then the configuration space of the whole system is $Voplus W$ (in the outer notion). Its elements are the pairs of vectors $(vec{r}_1,vec{r}_2)$, where $vec{r}_1$ is the position of the first particle while $vec{r}_2$ the position of the second. This is of course isomorphic to $mathbb{R}^6$, in which the first three elements of $(x_1,y_1,z_1,x_2,y_2,z_2)$ are interpreted to be the position of the first particle while the second three the position of the second.
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
$endgroup$
add a comment |
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$begingroup$
The x-y plane is of dimension 2. While each line is of dimension 1. 1+1=2.
Let me tell you a bit more about direct sums. Given a vector space $U$ and two subspaces $V$ and $W$, we say $U$ is the direct sum of $V$ and $W$ (and write $U=Voplus W$) if the only vector which $V$ and $W$ share is $0$, i.e. $Vcap W={0}$, and every vector in $U$ can be written as a sum of a vector in $V$ and a vector in $W$, i.e. $U=V+W$. This notion of direct sum is called inner direct sum. It is very visual and easy to understand. For example, consider the $x$-axis $V:=text{span}{(1,0)}={(x,0)inmathbb{R}^2}$ and the $y$-axis $W:=text{span}{(0,1)}={(0,y)inmathbb{R}^2}$ in the plane. We then quickly check $U:=mathbb{R}^2=Voplus W$. Notice however that, unless we add more structure, the direct sum does not encode the orthogonality of the $x$ and $y$ axes. Indeed, if one considers the diagonal $Z:=text{span}{(1,1)}={(x,x)inmathbb{R}^2}$ we also have $U=Voplus Z$ and $U=Zoplus W$.
Now, suppose you have two (unrelated as far as we care) vector spaces $V$ and $W$. We may ask if there is some bigger vector space $U$ which contains $V$ and $W$ and $U=Voplus W$. This is always possible and the construction is called the outer direct sum. We define the space $Voplus W$ as the set of all tuples $(v,w)$ where $vin V$ and $win W$. This forms a vector space with the addition $(v_1,w_1)+(v_2,w_2):=(v_1+v_2,w_1+w_2)$ and scalar multiplication $k(v,w):=(kv,kw)$. We may then consider the spaces $tilde{V}:={(v,0)in Uoplus W}$ and $tilde{W}:={(0,w)in Uoplus W}$. $V$ and $W$ are isomorphic to $tilde{V}$ and $tilde{W}$ as vector spaces respectively. Both are subspaces of $Voplus W$ and in fact $Voplus W=tilde{V}oplustilde{W}$. In the left of this equation we have the outer direct sum just defined while in the second we have the inner direct sum defined in the previous paragraph. It is thus costumary to identify $V$ and $W$ with $tilde{V}$ and $tilde{W}$ repectively. Moreover, we use the notation $v+w:=(v,w)$ for all $vin V$ and $win W$. Then both notions of inner product are virtually undistinguishable and most authors don't care to distinguish between them. Notice that with the outer notion of direct sum we truly have $mathbb{R}^2=mathbb{R}oplusmathbb{R}$.
In physics the direct sum pops up whenever vector spaces do. In classical mechanics, for example, the position of a particle is described by the vector space (well, not really but let us avoid the problem of an origin for the moment) $V=mathbb{R}^3$. Now consider we have a second particle whose position is described by the vector space $W=mathbb{R}^3$. Then the configuration space of the whole system is $Voplus W$ (in the outer notion). Its elements are the pairs of vectors $(vec{r}_1,vec{r}_2)$, where $vec{r}_1$ is the position of the first particle while $vec{r}_2$ the position of the second. This is of course isomorphic to $mathbb{R}^6$, in which the first three elements of $(x_1,y_1,z_1,x_2,y_2,z_2)$ are interpreted to be the position of the first particle while the second three the position of the second.
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
$endgroup$
add a comment |
$begingroup$
The x-y plane is of dimension 2. While each line is of dimension 1. 1+1=2.
Let me tell you a bit more about direct sums. Given a vector space $U$ and two subspaces $V$ and $W$, we say $U$ is the direct sum of $V$ and $W$ (and write $U=Voplus W$) if the only vector which $V$ and $W$ share is $0$, i.e. $Vcap W={0}$, and every vector in $U$ can be written as a sum of a vector in $V$ and a vector in $W$, i.e. $U=V+W$. This notion of direct sum is called inner direct sum. It is very visual and easy to understand. For example, consider the $x$-axis $V:=text{span}{(1,0)}={(x,0)inmathbb{R}^2}$ and the $y$-axis $W:=text{span}{(0,1)}={(0,y)inmathbb{R}^2}$ in the plane. We then quickly check $U:=mathbb{R}^2=Voplus W$. Notice however that, unless we add more structure, the direct sum does not encode the orthogonality of the $x$ and $y$ axes. Indeed, if one considers the diagonal $Z:=text{span}{(1,1)}={(x,x)inmathbb{R}^2}$ we also have $U=Voplus Z$ and $U=Zoplus W$.
Now, suppose you have two (unrelated as far as we care) vector spaces $V$ and $W$. We may ask if there is some bigger vector space $U$ which contains $V$ and $W$ and $U=Voplus W$. This is always possible and the construction is called the outer direct sum. We define the space $Voplus W$ as the set of all tuples $(v,w)$ where $vin V$ and $win W$. This forms a vector space with the addition $(v_1,w_1)+(v_2,w_2):=(v_1+v_2,w_1+w_2)$ and scalar multiplication $k(v,w):=(kv,kw)$. We may then consider the spaces $tilde{V}:={(v,0)in Uoplus W}$ and $tilde{W}:={(0,w)in Uoplus W}$. $V$ and $W$ are isomorphic to $tilde{V}$ and $tilde{W}$ as vector spaces respectively. Both are subspaces of $Voplus W$ and in fact $Voplus W=tilde{V}oplustilde{W}$. In the left of this equation we have the outer direct sum just defined while in the second we have the inner direct sum defined in the previous paragraph. It is thus costumary to identify $V$ and $W$ with $tilde{V}$ and $tilde{W}$ repectively. Moreover, we use the notation $v+w:=(v,w)$ for all $vin V$ and $win W$. Then both notions of inner product are virtually undistinguishable and most authors don't care to distinguish between them. Notice that with the outer notion of direct sum we truly have $mathbb{R}^2=mathbb{R}oplusmathbb{R}$.
In physics the direct sum pops up whenever vector spaces do. In classical mechanics, for example, the position of a particle is described by the vector space (well, not really but let us avoid the problem of an origin for the moment) $V=mathbb{R}^3$. Now consider we have a second particle whose position is described by the vector space $W=mathbb{R}^3$. Then the configuration space of the whole system is $Voplus W$ (in the outer notion). Its elements are the pairs of vectors $(vec{r}_1,vec{r}_2)$, where $vec{r}_1$ is the position of the first particle while $vec{r}_2$ the position of the second. This is of course isomorphic to $mathbb{R}^6$, in which the first three elements of $(x_1,y_1,z_1,x_2,y_2,z_2)$ are interpreted to be the position of the first particle while the second three the position of the second.
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
$endgroup$
add a comment |
$begingroup$
The x-y plane is of dimension 2. While each line is of dimension 1. 1+1=2.
Let me tell you a bit more about direct sums. Given a vector space $U$ and two subspaces $V$ and $W$, we say $U$ is the direct sum of $V$ and $W$ (and write $U=Voplus W$) if the only vector which $V$ and $W$ share is $0$, i.e. $Vcap W={0}$, and every vector in $U$ can be written as a sum of a vector in $V$ and a vector in $W$, i.e. $U=V+W$. This notion of direct sum is called inner direct sum. It is very visual and easy to understand. For example, consider the $x$-axis $V:=text{span}{(1,0)}={(x,0)inmathbb{R}^2}$ and the $y$-axis $W:=text{span}{(0,1)}={(0,y)inmathbb{R}^2}$ in the plane. We then quickly check $U:=mathbb{R}^2=Voplus W$. Notice however that, unless we add more structure, the direct sum does not encode the orthogonality of the $x$ and $y$ axes. Indeed, if one considers the diagonal $Z:=text{span}{(1,1)}={(x,x)inmathbb{R}^2}$ we also have $U=Voplus Z$ and $U=Zoplus W$.
Now, suppose you have two (unrelated as far as we care) vector spaces $V$ and $W$. We may ask if there is some bigger vector space $U$ which contains $V$ and $W$ and $U=Voplus W$. This is always possible and the construction is called the outer direct sum. We define the space $Voplus W$ as the set of all tuples $(v,w)$ where $vin V$ and $win W$. This forms a vector space with the addition $(v_1,w_1)+(v_2,w_2):=(v_1+v_2,w_1+w_2)$ and scalar multiplication $k(v,w):=(kv,kw)$. We may then consider the spaces $tilde{V}:={(v,0)in Uoplus W}$ and $tilde{W}:={(0,w)in Uoplus W}$. $V$ and $W$ are isomorphic to $tilde{V}$ and $tilde{W}$ as vector spaces respectively. Both are subspaces of $Voplus W$ and in fact $Voplus W=tilde{V}oplustilde{W}$. In the left of this equation we have the outer direct sum just defined while in the second we have the inner direct sum defined in the previous paragraph. It is thus costumary to identify $V$ and $W$ with $tilde{V}$ and $tilde{W}$ repectively. Moreover, we use the notation $v+w:=(v,w)$ for all $vin V$ and $win W$. Then both notions of inner product are virtually undistinguishable and most authors don't care to distinguish between them. Notice that with the outer notion of direct sum we truly have $mathbb{R}^2=mathbb{R}oplusmathbb{R}$.
In physics the direct sum pops up whenever vector spaces do. In classical mechanics, for example, the position of a particle is described by the vector space (well, not really but let us avoid the problem of an origin for the moment) $V=mathbb{R}^3$. Now consider we have a second particle whose position is described by the vector space $W=mathbb{R}^3$. Then the configuration space of the whole system is $Voplus W$ (in the outer notion). Its elements are the pairs of vectors $(vec{r}_1,vec{r}_2)$, where $vec{r}_1$ is the position of the first particle while $vec{r}_2$ the position of the second. This is of course isomorphic to $mathbb{R}^6$, in which the first three elements of $(x_1,y_1,z_1,x_2,y_2,z_2)$ are interpreted to be the position of the first particle while the second three the position of the second.
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
$endgroup$
The x-y plane is of dimension 2. While each line is of dimension 1. 1+1=2.
Let me tell you a bit more about direct sums. Given a vector space $U$ and two subspaces $V$ and $W$, we say $U$ is the direct sum of $V$ and $W$ (and write $U=Voplus W$) if the only vector which $V$ and $W$ share is $0$, i.e. $Vcap W={0}$, and every vector in $U$ can be written as a sum of a vector in $V$ and a vector in $W$, i.e. $U=V+W$. This notion of direct sum is called inner direct sum. It is very visual and easy to understand. For example, consider the $x$-axis $V:=text{span}{(1,0)}={(x,0)inmathbb{R}^2}$ and the $y$-axis $W:=text{span}{(0,1)}={(0,y)inmathbb{R}^2}$ in the plane. We then quickly check $U:=mathbb{R}^2=Voplus W$. Notice however that, unless we add more structure, the direct sum does not encode the orthogonality of the $x$ and $y$ axes. Indeed, if one considers the diagonal $Z:=text{span}{(1,1)}={(x,x)inmathbb{R}^2}$ we also have $U=Voplus Z$ and $U=Zoplus W$.
Now, suppose you have two (unrelated as far as we care) vector spaces $V$ and $W$. We may ask if there is some bigger vector space $U$ which contains $V$ and $W$ and $U=Voplus W$. This is always possible and the construction is called the outer direct sum. We define the space $Voplus W$ as the set of all tuples $(v,w)$ where $vin V$ and $win W$. This forms a vector space with the addition $(v_1,w_1)+(v_2,w_2):=(v_1+v_2,w_1+w_2)$ and scalar multiplication $k(v,w):=(kv,kw)$. We may then consider the spaces $tilde{V}:={(v,0)in Uoplus W}$ and $tilde{W}:={(0,w)in Uoplus W}$. $V$ and $W$ are isomorphic to $tilde{V}$ and $tilde{W}$ as vector spaces respectively. Both are subspaces of $Voplus W$ and in fact $Voplus W=tilde{V}oplustilde{W}$. In the left of this equation we have the outer direct sum just defined while in the second we have the inner direct sum defined in the previous paragraph. It is thus costumary to identify $V$ and $W$ with $tilde{V}$ and $tilde{W}$ repectively. Moreover, we use the notation $v+w:=(v,w)$ for all $vin V$ and $win W$. Then both notions of inner product are virtually undistinguishable and most authors don't care to distinguish between them. Notice that with the outer notion of direct sum we truly have $mathbb{R}^2=mathbb{R}oplusmathbb{R}$.
In physics the direct sum pops up whenever vector spaces do. In classical mechanics, for example, the position of a particle is described by the vector space (well, not really but let us avoid the problem of an origin for the moment) $V=mathbb{R}^3$. Now consider we have a second particle whose position is described by the vector space $W=mathbb{R}^3$. Then the configuration space of the whole system is $Voplus W$ (in the outer notion). Its elements are the pairs of vectors $(vec{r}_1,vec{r}_2)$, where $vec{r}_1$ is the position of the first particle while $vec{r}_2$ the position of the second. This is of course isomorphic to $mathbb{R}^6$, in which the first three elements of $(x_1,y_1,z_1,x_2,y_2,z_2)$ are interpreted to be the position of the first particle while the second three the position of the second.
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
answered Jan 28 at 17:04
Iván Mauricio BurbanoIván Mauricio Burbano
37218
37218
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4
$begingroup$
If you are adding dimensions then you get $1+1=2$. I don't think I understand the issue. So you go from two spaces on $mathbb{R}$ (each dimension $1$) to the space on $mathbb{R}^2$ (dimension $2$).
$endgroup$
– Aaron Stevens
Jan 28 at 17:11
1
$begingroup$
Would Mathematics be a better home for this question.
$endgroup$
– Qmechanic
Jan 28 at 17:13